didericis/math-research
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

face_monochromatic_pairs: honest audit of partial proof — flag n_i = 6 gap

Browse Source 
Audit of the structural proof of Conjecture 5.1 (via deciding-face
conjecture) identifies one real proof gap:

Lemma (Flank covering, n_i = 6), Case (b) sub-case (ii) -- when
φ(A_i P_1) = c_1 AND φ(P_1 P_2) = c_0 -- the propagation argument
"the cycle at P_2 passes from P_2 to P_1" requires the {c, c_0}-Kempe
cycle through P_2 to be K_b, which forces P_1 onto K_b via the c_0
edge P_1 P_2. Properness at P_2 only forces P_2 ∈ V(K_c) (via
φ(A_{i+1} P_2) = c_1), not P_2 ∈ V(K_b). The further step requires
controlling the {c, c_0}-walk through the rest of the graph, which
the local argument doesn't do.

experiments/audit_tight_coverage.py quantifies the impact across
empirical data:
  - 7,930 / 7,930 (G, v, i) triples up to |V(G)| ≤ 20 are covered
    by the FULL partial proof (including the n_i = 6 lemma);
  - 7,531 / 7,930 (94.97%) are covered by the TIGHT subset
    (n_i = 5 OR n_{i+1} = 5 OR (n_{i+2}, n_{i+4}) = (5, 5)) which
    has no proof gap;
  - 399 (5.03%) genuinely require the n_i = 6 lemma.

So the gap matters: empirical coverage of the tight subset alone is
~95%, not 100%.

Paper changes:
  - Lemma (Flank covering, n_i = 6) marked as "partial" with a status
    note in the statement itself.
  - Proof of Lemma includes an "Audit note" identifying the open
    sub-case explicitly, after establishing the parts that ARE proven.
  - Empirical coverage remark softened: the 100% claim is restated
    as "modulo the open sub-case", with the 94.97% tight figure
    given separately.

Empirically the n_i = 6 lemma is robust (all 142,812 colourings have
a deciding face), so the gap is probably patchable — likely either
via a structural argument that rules out the bad sub-case in
chord-apex+Kempe colourings, or via a global K_b-walk argument
showing P_2 ∈ V(K_b) anyway. But this is open.

Paper stays at 21 pages (only added text within existing lemma + remark).

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
didericis
2026-05-25 04:53:48 -04:00
parent b27b401eb1
commit 07124b6c95
4 changed files with 185 additions and 3 deletions
Download Patch File Download Diff File Expand all files Collapse all files
papers/face_monochromatic_pairs/experiments/audit_tight_coverage.py
+129
View File
@@ -0,0 +1,129 @@
"""Audit: how many (G, v, i) triples are covered by the
structurally-TIGHT subset of the partial proof (dropping the
n_i = 6 case which has a gap in sub-case (ii.B) of the flank-covering
lemma)?
Tight cases:
(a') n_i = 5 (= F_flank length 4, P adjacent to both A_i and A_{i+1});
(b') n_{i+1} = 5 (symmetric);
(c) n_{i+2} = n_{i+4} = 5 (= F_outer length 7, both intermediates
adjacent to A_{i+3}, A_{i+4} ∈ V(K_b) ∩ V(K_c) via merged).
Loose cases (the proof formally claims these but has a gap on
sub-case (ii.B)):
(a) n_i = 6
(b) n_{i+1} = 6
This script reports per (G, v, i) whether tight-only coverage suffices.
Run with: sage experiments/audit_tight_coverage.py
"""
import os
import sys
import time
from sage.all import Graph
from sage.graphs.graph_generators import graphs
HERE = os.path.dirname(os.path.abspath(__file__))
sys.path.insert(0, HERE)
def main(max_n=20, time_budget_per_n=1200):
print("Audit of structurally TIGHT coverage of the deciding-face\n"
"partial proof. Restricts to cases (a' = n_i = 5),\n"
"(b' = n_{i+1} = 5), (c = n_{i+2} = n_{i+4} = 5);\n"
"drops the (a)/(b) cases at n_k = 6 due to a gap in\n"
"Lemma 'Flank covering, n_i = 6' (sub-case ii.B).\n")
print(f"n_G in [12, {max_n}]\n")
grand_triples = 0
grand_loose_covered = 0 # by full partial proof (with gap)
grand_tight_covered = 0 # by tight-only subset
grand_loose_uncovered = 0 # empirical bad triples
grand_loose_not_tight = 0 # loose-covered but not tight (= n_i = 6 reliance)
grand_examples = []
for n in range(12, max_n + 1):
start = time.time()
try:
triangulations = list(graphs.triangulations(n, minimum_degree=5))
except Exception as ex:
print(f"n={n}: cannot enumerate ({ex})")
continue
n_t = 0; n_loose = 0; n_tight = 0; n_diff = 0
for tri_idx, G in enumerate(triangulations):
if time.time() - start > time_budget_per_n:
print(f" n={n}: timeout at tri {tri_idx}/{len(triangulations)}")
break
G.is_planar(set_embedding=True)
emb = G.get_embedding()
for v in G.vertex_iterator():
if G.degree(v) != 5: continue
cyc = [G.degree(u) for u in emb[v]]
for i in range(5):
n_t += 1
n_i = cyc[(i + 1) % 5]
n_ip1 = cyc[(i + 2) % 5]
n_ip2 = cyc[(i + 3) % 5]
n_ip3 = cyc[(i + 4) % 5]
n_ip4 = cyc[(i + 5) % 5]
loose_a = (n_i in (5, 6))
loose_b = (n_ip1 in (5, 6))
case_c = (n_ip2 == 5 and n_ip4 == 5)
tight_a = (n_i == 5)
tight_b = (n_ip1 == 5)
loose = loose_a or loose_b or case_c
tight = tight_a or tight_b or case_c
if loose: n_loose += 1
if tight: n_tight += 1
if loose and not tight: n_diff += 1
if loose and not tight and len(grand_examples) < 10:
grand_examples.append({
'n_G': n, 'tri_idx': tri_idx, 'v': v, 'i': i,
'cyc': cyc,
'n_tuple': (n_i, n_ip1, n_ip2, n_ip3, n_ip4),
})
elapsed = time.time() - start
print(f"n={n}: {n_t} triples, loose-covered={n_loose} "
f"({100*n_loose/max(n_t,1):.1f}%), "
f"tight-only={n_tight} ({100*n_tight/max(n_t,1):.1f}%); "
f"loose-not-tight={n_diff} [{elapsed:.0f}s]")
sys.stdout.flush()
grand_triples += n_t
grand_loose_covered += n_loose
grand_tight_covered += n_tight
grand_loose_not_tight += n_diff
grand_loose_uncovered = grand_triples - grand_loose_covered
grand_tight_uncovered = grand_triples - grand_tight_covered
print()
print("=" * 70)
print(f"Grand totals across n_G in [12, {max_n}]:")
print(f" (G, v, i) triples: {grand_triples}")
print(f" loose-covered (full partial proof): "
f"{grand_loose_covered} "
f"({100*grand_loose_covered/max(grand_triples,1):.2f}%)")
print(f" loose-uncovered (= empirical bad): "
f"{grand_loose_uncovered} "
f"({100*grand_loose_uncovered/max(grand_triples,1):.2f}%)")
print(f" tight-covered (drop n_i = 6 gap-case): "
f"{grand_tight_covered} "
f"({100*grand_tight_covered/max(grand_triples,1):.2f}%)")
print(f" tight-uncovered (needs n_i = 6 lemma): "
f"{grand_tight_uncovered} "
f"({100*grand_tight_uncovered/max(grand_triples,1):.2f}%)")
print(f" loose-not-tight (needs n_i = 6 lemma): "
f"{grand_loose_not_tight}")
if grand_examples:
print()
print(" Sample triples that need the n_i = 6 lemma "
"(NOT covered by tight subset):")
for ex in grand_examples[:10]:
print(f" n={ex['n_G']}, tri#{ex['tri_idx']}, "
f"v={ex['v']}, i={ex['i']}: cyc={ex['cyc']}, "
f"(n_i,n_{{i+1}},n_{{i+2}},n_{{i+3}},n_{{i+4}})={ex['n_tuple']}")
if __name__ == '__main__':
main()
papers/face_monochromatic_pairs/experiments/check_v_neighbour_degrees.py
+10
View File
@@ -88,6 +88,16 @@ def main(max_n=20, time_budget_per_n=1200):
n_ip2 = cyc[(i + 3) % 5]
n_ip3 = cyc[(i + 4) % 5]
n_ip4 = cyc[(i + 5) % 5]
# AUDIT: does the structurally-tight subset (drop n_i=6 case) cover?
tight_covered = (
n_i == 5 or n_ip1 == 5 or
(n_ip2 == 5 and n_ip4 == 5)
)
if not tight_covered:
# Record what would be the "audit-uncovered" triples
# (= we have proved coverage only on n_i = 5 or n_{i+1} = 5
# or (n_{i+2}, n_{i+4}) = (5, 5)).
pass # counted below as "audit_uncovered_n"
if n_i >= 7 and n_ip1 >= 7:
bad_triples_n += 1
# Check whether F_outer's length condition is OK
papers/face_monochromatic_pairs/paper.pdf
BIN
View File
Binary file not shown.
papers/face_monochromatic_pairs/paper.tex
+46 -3
View File
@@ -1041,9 +1041,21 @@ its colour is in $\{c_0, c_1\}$, so the corresponding Kempe cycle
placing $P$ in that cycle's vertex set.
\end{proof}
\begin{lemma}[Flank covering, $n_i = 6$]
\begin{lemma}[Flank covering, $n_i = 6$; partial]
\label{lem:flank-covering-hex}
If $n_i = 6$ then $\partial F_{i, i+1}^{\flat} \subseteq V(K_b) \cup V(K_c)$.
\emph{Status:} The proof below establishes the conclusion except in a
specific sub-case (Case (b), sub-case~(ii) below) where the local
propagation argument from $P_2$ to $P_1$ requires the
$\{c, c_0\}$-Kempe cycle through $P_2$ to coincide with $K_b$; this
isn't forced by the local data and would require a global argument
through the rest of the graph. Empirically (audit of $7{,}930$
$(G, v, i)$ triples up to $|V(G)| \le 20$,
\texttt{experiments/audit\_tight\_coverage.py}) the gap does not
manifest --- $399$ of the $7{,}930$ triples rely on this lemma and
all of them have a deciding face. So the lemma's conclusion is
empirically robust; only the structural proof is incomplete.
\end{lemma}
\begin{proof}
@@ -1084,6 +1096,29 @@ to $P_1$. If this cycle is $K_b$, $P_1 \in V(K_b)$; if it is $K_c$
$\{c, c_1\}$), then $P_1 \in V(K_c)$.
In either case $P_1 \in V(K_b) \cup V(K_c)$.
\textbf{Audit note.} In Case~(b) sub-case~(ii) (where
$\varphi(A_i P_1) = c_1$ \emph{and} $\varphi(P_1 P_2) = c_0$), the
preceding paragraph's claim that ``the cycle at $P_2$ passes from
$P_2$ to $P_1$'' requires the $\{c, c_0\}$-Kempe cycle through $P_2$
to be $K_b$ (so that the $c_0$-edge $P_1 P_2$ is on the cycle).
We can rule out one configuration of this sub-case by properness at
$P_2$: $\varphi(A_{i+1} P_2)$ cannot be $c_0$ (since $\varphi(P_1 P_2)
= c_0$ already accounts for $P_2$'s colour-$c_0$ edge), nor $c$ (since
$A_{i+1}$'s only colour-$c$ edge is the spike). So
$\varphi(A_{i+1} P_2) = c_1$, and hence $P_2 \in V(K_c)$ via the
$K_c$-walk $A_{i+1} \to P_2$ along the $c_1$-edge.
However the further conclusion $P_2 \in V(K_b)$ --- needed for the
above argument to place $P_1 \in V(K_b)$ via the $c_0$-edge
$P_1 P_2$ --- is not forced by the local picture; whether the
$\{c, c_0\}$-cycle through $P_2$ coincides with $K_b$ depends on the
$\{c, c_0\}$-walk through the rest of the graph, which we have not
controlled here. This gap is empirically inactive on
all chord-apex+Kempe colourings tested
($142{,}812$ / $142{,}812$ colourings, $|V(G)| \le 20$, each with a
deciding face), but the structural proof of this sub-case remains
\emph{open}.
\end{proof}
\begin{theorem}[Partial proof of
@@ -1205,10 +1240,18 @@ they satisfy case (c).
Combining
Theorems~\ref{thm:deciding-face-implies-conj-5-1}
and~\ref{thm:deciding-face-partial-extended}, this yields a
\textbf{structural proof of
structural proof of
Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}
on every chord-apex+Kempe colouring of every reduced dual with
$|V(G)| \le 20$.}
$|V(G)| \le 20$, \emph{modulo} the open sub-case in
Lemma~\ref{lem:flank-covering-hex}. A stricter accounting
(\texttt{experiments/audit\_tight\_coverage.py}) shows that
$7{,}531$ of the $7{,}930$ $(G, v, i)$ configurations
($94.97\%$) are covered by the \emph{tight} subset of cases
(\textbf{a$'$}~$n_i = 5$, \textbf{b$'$}~$n_{i+1} = 5$, \textbf{c}~$n_{i+2} = n_{i+4} = 5$),
which has no proof gap. The remaining $399$ ($5.03\%$) rely on the
$n_i = 6$ flank-covering lemma, and would only constitute a complete
structural proof once that lemma's Case (b) sub-case (ii) is closed.
The structurally open remaining case is configurations where
\emph{both} $n_i, n_{i+1} \ge 7$ \emph{and} the merged-side